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andrew-mc [135]
1 year ago
7

a sample of a metal, heated in air, gains mass dm by oxidising at a rate that declines with time t in a parabolic way, meaning t

hat dm2 ¼ kpt where kp is a constant. why does the rate fall off in this way? what does it imply about the nature and utility of the oxide?
Chemistry
1 answer:
velikii [3]1 year ago
5 0

Oxidation originates at the surface of the metal and then it slowly penetrates through the surface.

<em>As at the beginning exposed surface area is more therefore rate of oxidation is also more. After that, the process needed to occur inside the exposed surface of the metal, and the rate of diffusion decreases, so the rate of oxidation decreases.</em>

<em />

<em>It represents oxide layers that reduce the diffusion rate, therefore it is used for producing isolation shields for metals. Apart from that for covering a metal from the external environment or masking oxide layers are produced on metal surfaces.</em>

<em />

<em />

Mass is a quantitative measure of inertia, an essential property of all dependents. It's far, in effect, the resistance that a frame of count number gives to a change in its speed or position upon the application of pressure. The more the mass of a frame, the smaller the alternate produced with the aid of a carried-out force.

Learn more about Mass here:-brainly.com/question/28021242

#SPJ4

<em />

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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
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<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

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How many metals, nonmetals and metalloids are there in the periodic table?
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Read 2 more answers
I need this please my brain hurts
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8.4 grams. I think but I’m not 100% sure
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